If for some $\alpha$ and $\beta$ in $\mathbb{R},$ the intersection of the following three planes $x+4y-2z=1$,$x+7y-5z=\beta$,and $x+5y+\alpha z=5$ is a line in $\mathbb{R}^{3},$ then $\alpha+\beta$ is equal to

The plane $ax + by = 0$ is rotated about its line of intersection with the plane $z = 0$ through an angle $\alpha$. Prove that the equation of the plane in its new position is $ax + by \pm (\sqrt{a^{2} + b^{2}} \tan \alpha) z = 0$.

$A$ plane $\pi$ passing through the point $3 \hat{i}-4 \hat{j}+5 \hat{k}$ is parallel to the plane which passes through the point $\hat{i}+\hat{j}-\hat{k}$ and is perpendicular to the vector $\hat{i}+2 \hat{j}-3 \hat{k}$. Then the Cartesian equation of $\pi$ is

$(1, -2, 1)$ is a point on a plane $\pi$ and $\pi$ is parallel to the plane $x-y-z=0$. If the equation of $\pi$ is $ax+by+cz-2=0$,then $b-2c=$

If the foot of the perpendicular from $(0,0,0)$ to a plane is $(1,2,3)$,then the equation of the plane is

If $M$ is the foot of the perpendicular drawn from $P(1,2,-1)$ to the plane passing through the point $A(3,-2,1)$ and perpendicular to the vector $4 \hat{i}+7 \hat{j}-4 \hat{k}$,then the length of $PM$ is